Obvious Rick and Morty reference aside, watching the episode make me curious: Can we make something that's level, even to the molecular scale? If so, how, and what would it feel like, to step on something that's flatter than the natural curvature of the earth?

Silicon wafers (not edible) are quite level, otherwise you'd have a really hard time etching the chips (also not edible). Still, they usually have bumps in the order of micrometres, so that's not really flat "on the molecular level". This piece of glass has bumps in the order of nanometres, so that's better (also it's 10cm wide, unlike pieces of graphene, which are truly flat but really tiny).

It's keel might be a straight line, so the water level in the middle might be one inch higher than the ends. I don't think that would be noticable, but I'm sure it had to be taken into account when building the thing.

Of course large structures (bridges, dams) are routinely built taking into account the curvature of the Earth.

andykhang wrote:Obvious Rick and Morty reference aside, watching the episode make me curious: Can we make something that's level, even to the molecular scale? If so, how, and what would it feel like, to step on something that's flatter than the natural curvature of the earth?

Hmmm, if gravity is quantized then I suppose you could. Otherwise, well I guess even then it's definite. Hmmm, but since you can never get to absolute zero you'd have to account for the individual gravitational effects of every possible moving molecule and particle in the vicinity, and somehow get them to cancel out to get to perfectly level. Could you do that?

Or is taking the graviational effects of every possible particle in your horizon going too far?

p1t1o wrote:I want to roll it across the worlds flattest surface. This is an experiment that needs to be done.

Of course if one attempts such an experiment in any significant gravity field the "worlds flattest surface" is not so likely to remain the worlds flattest, and if done in microgravity "roll" become somewhat meaningless. https://xkcd.com/669/

e.g. 9.8N (say 1kg at typical earth gravitational acceleration) / 0m2 = "Math Error" ???! I need to get a better calculator!, but anyway it's something like (reallybignumber)YPa and I'm sure it is enough pressure to dent diamond, Sure once you put a dent in your worlds flattest surface the contact area becomes a lot bigger than 0, but a "flat surface" with a dent in it is not so very "flat". Well, okay, it might still be the "flattest" (with the smallest dent) at that particular time, but it won't be as flat as before you rolled the ball across it. Which means it is possible for there to be a flatter surface in existence that has not had a ball rolled across it!

If the contact area were truly one atom, or about (210 pm)2 = 44,000 pm2. then the pressure of the 1 kg ball in 1 standard gravity would be about 2.2 × 1020 Pa = 220 EPa. By comparison, at least some (microscopic) indentation should be made on a diamond surface at something like 50-100 GPa, so the pressure would be easily billions of times what would be required to deform even diamond.

However, the ball is not made of diamond; it is made of silicon. So if the surface is much harder than silicon, it is the ball that should deform, and the surface could still remain level. Moreover, the deformation should be nearly elastic, so the rolling motion should still have very little resistance.

Of course, in reality, the ball of silicon is not atomically perfect anyway, and neither is the surface. He never said he wanted an atomically perfect ball rolled on an atomically perfect level surface, just the best the world has to offer. There is nothing preventing us from performing that experiment, except the huge waste it would incur.

Well, I was going with the Van der Waals radius, which seemed like a reasonable estimate for a single-atom contact. But yeah, in reality there will obviously be many atoms in contact. You will also, in reality, get some deformation at any nonzero pressure. There isn't really any such thing as a "perfect sphere."

In Kerbal Space Program, the planet Kerbal is round while the runway is flat, which means that the center of the runway is actually lower than the ends. This creates an odd effect where it is most fuel efficient method of take off is to stay on the runway as long as possible, without going past the half-way mark.

This is unrelated to the discussion, but I felt a compelling need to bring it up anyway.

"You are not running off with Cow-Skull Man Dracula Skeletor!" -Socrates

That's just one proposed standard. The 2018 redefinition is likely to be based on a watt balance. Currently, the international prototype is still the official definition.

It's still a really cool experiment, though. They claim to have measured Avogadro's constant to nine digits of precision.

If the sphere is used for the 2018 redefinition, they will have Avogadro's constant to infinite precision (although I wouldn't claim they "measured" such).

I'd look to Saturn's rings for examples of extreme flatness. Silicon wafers might be atomically flat, but they max out around 300mm to 450mm in diameter. Saturn's rings are *big* and still amazingly flat (and thin).

How are we defining level, anyway? This is presumably a technical discussion, but nobody seems to have hammered down how to check "levelness". I would define level as having the property that the "average surface" is parallel with the gravitational field. The "average surface" is then defined as a plane[?] having the least squares fit to all parts of the top of the surface [assumes that only the top matters, for things like picture hanging, then average over the front surface and make it orthogonal to the gravity field].

This leaves the question as to whether a flat plane can be as level as a curved arc matching the gravitational field. It is a bit of a philosophical/language issue, but I will point out that the [fully upgraded] landing strip used in Kerbal Space Program is absolutely flat, and this means that the elevation is higher at the start and end of the runway than in the middle, thanks to the curvature of Kerbin [which is 1/10th the size of Earth].